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Math200b, Lecture 11 Math200b, lecture 11 Golsefidy We have proved that a module is projective if and only if it is a direct summand of a free module; in particular, any free module is projective. For a given ring A, we would like to know to what extent the converse of this statement holds; and if it fails, we would like to somehow “measure" how much it does! In genral this is hard question; in your HW assignement you will show that for a local commutative ring A, any finitely generated projective module is free. By a result of Kaplansky the same statement holds for a module that is not necessarily finitely generated. Next we show that for a PID, any finitely generated projective module is free. Proposition 1 Let D be an integral domain. Then free D-module ) Projective ) torsion-free. 1 If D is a PID, for a finitely generated D-module all the above proper- ties are equivalent. Proof. We have already discussed that a free module is projec- tive. A projective module is a direct summand of a free module and a free module of an integral domain is torsion free. By the fundamental theorem of finitely generated modules over a PID, a torsion free finitely generated module over D is free; and claim follows. p Next we show A » −10¼ has a finitely generated projec- tive module that is not free. In fact any ideal of A is projective; and since it is not a PID, it has an ideal that is not free. Based on the mentioned result of Kaplansky, a projective module is locally free. And for finitely generated modules, the converse of this statement holds as well: a finitely generated locally free module is projective. Hence by the previous proposition, if A is a Noetherian integral domain and Ap is a PID for any p 2 Spec¹Aº A , then anyp ideal of is projective. In math200c, we will prove that » −10¼ has this property (it is a Dedekind domain). For now, however, we present a hands-on approach and point out the connection with fractional ideals, having an 2 inverse as a fractional ideal, and being projective. Lemma 2 Suppose D is an integral domian and a E D. Then a is a free D-module if and only if a is a princpal ideal. Proof. ()) Since a is a submodule of D, ranka ≤ rankD 1. So either rank a 0 or rank a 1. Since D has no zero-divisors, rank a 0 implies a 0. Hence if a non-zero ideal is a free D-module, then it has rank 1; and so a aD for some a 2 D. And claim follows. (() Since a is principal, a aD for some a 2 D. If a 0, then a 0 is a free D-module. If a , 0, then x 7! ax is a D- module isomorphism from D to a; and so a is a free D-module. p p Example. Let A » −10¼ and a : h2, −10i. Then a is not a free A-module. Proof. Suppose to the contrary that a is a free A-module; a a Aa then by the previous lemmap is a principal ideal. So for some a. So aj2 and aj −10; hence N¹aºj gcd¹4, 10º 2. So N¹aº is either 1 or 2. Since there is no x, y 2 such that x2 + 10y2 2, we deduce that N¹aº 1, which implies a is a 3 unit; and so a A. On the other hand, one can see that, for any z 2 a, N¹zº is even; and so a is a proper ideal. Overall we get that a is not a free A-module. In the next lecture, we show that a is a projective A-module. 4.
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